Table Of ContentASSOCIATED PRIMES OF THE SQUARE OF THE ALEXANDER
DUAL OF HYPERGRAPHS
9 ASHOKCUTKOSKY
0
0
2 Abstract. The purpose of this paper is to provide methods for determining the as-
sociated primes of (I(H)∨)2 for an m-hypergraph H. We prove a general method for
n
detecting associated primes of the square of the Alexander dual of the edge ideal based
a
J oncombinatorialconditionsonthem-hypergraph. Also,wedemonstrateamoreefficient
combinatorialcriterionfordetectingthenon-existenceofnon-minimalassociatedprimes.
2
1 Ininvestigating3-hypergraphs,weproveasurprisingextensionofthepreviouslydiscov-
ered results for 2-hypergraphs (simple graphs). For 2-hypergraphs, associated primes
] of the square of the Alexander dual of the edge ideal are either of height 2 or of odd
C
heightgreaterthan2. However,weprovethatinthe3-hypergraphcase,thereisnosuch
A restriction -orindeedanyrestriction -ontheheightsof theassociated primes. Further,
. wegeneralizethisresulttoanydimensiongreaterthan3. Specifically,givenanyintegers
h m, q, and n with 3 ≤ m ≤ q ≤ n, we construct a m-hypergraph of size n with an
t
a associated primeofheightq. Wefurtherprovethatitispossibletoconstructconnected
m m-hypergraphsunderthesame conditions.
[
1
v
Introduction
8
7
6 An m-hypergraph is a vertex set {x1,...,xn} and an edge set
1
{{x ,...,x },{x ,...,x },...}
1. i1 im j1 jm
0 whereeachedgejoinsmvertices. Manycombinatorialpropertiesofhypergraphshavebeen
9
definedandstudied. However, itispossible,andoftenadvantageous, tostudyhypergraphs
0
from an algebraic rather than a combinatorial point of view. This is done by associating
:
v
to the vertex set a polynomial ring in n variables over a field and associating to the edge
i
X set an ideal, I, called the edge ideal in that ring. The edge ideal is the ideal generated
r by monomials x ···x where {x ,...,x } is an edge. There is a further function called
a i1 im i1 im
the Alexander dual that transforms the edge ideal of a hypergraph into another ideal, I∨,
obtained via intersection of ideals of the form (x ,...,x ) where {x ,...,x } is an edge.
i1 im i1 im
Thisnewidealhastheinterestingpropertythattheminimalgeneratorscorrespondexactly
to the minimal vertex covers of the hypergraph. We can also take symbolic powers of the
Alexander dual, which have slightly different properties. In particular, the symbolic kth
power of the Alexander dual is generated by the monomials corresponding to the k-covers
of the hypergraph, while the ordinary kth powers of the Alexander dual are generated by
the k-covers that are sums of 1-covers, a phenomenon that has been studied by Herzog,
Hibi and Trung [3], Gitler, Reyes and Villarreal [2], Fransisco, H`a and Van Tuyl [1] and
other mathematicians. Herzog, Hibi and Trung prove in [3] that the algebra of vertex
covers of a hypergraph (the symbolic algebra of the Alexander dual) is finitely generated.
It is proven in [1] that in the case of 2-hypergraphs (or simple graphs), the associated
primes of (I∨)2 correspond to the edges of the graph, and to the odd induced cycles.
We prove in Theorem 2.6 of this paper that for any integers n, q and m with n ≥ q ≥
m ≥ 3, there exists an m-hypergraph of any size n such that the square of the Alexander
1
dual of the edge ideal of the m-hypergraph has an associated prime of height q. We also
prove that for m-hypergraphs, it is not possible to have an associated prime of height
less than m. Further, we have a generalized statement (Theorem 1.6) for m-hypergraphs
that an associated prime for an induced subhypergraph is an associated prime for the full
hypergraph. Using the above statements, one can prove that for any dimension higher
than 2, there are connected hypergraphs of any size with any possible height of associated
prime. These results are in stark contrast with the behavior of 2-hypergraphs, where it
is not possible to obtain associated primes of even height greater than 2 (Theorem 1.1
[1]). Also, we have in Theorem 1.4 a necessary and sufficient combinatorial criterion for
determining the existence of and an explicit description for the associated primes. Briefly,
themethodinvolvesfindinga2-coverthatisnotthesumoftwo1-covers-thisrepresentsan
element notin(I(H)∨)2. Next, wenotewhich, if any, components itis possibleto increase
by 1 to make a 2-cover that is the sum of 1-covers, and prove that adding anything to any
other component does not make a sumof two 1-covers. Thenthe components noted in the
second step correspond to the generators of an associated prime. This method provides a
nice combinatorial gateway between the structure of the hypergraph and the structure of
the algebra.
While we have found numerous examples to show that the elements of a cycle do not
necessarily generate an associated prime, it is an interesting open question as to whether
the converse is true. Further, although in many examples the associated primes of higher
powers of the Alexander dual appear to be the same as for the second power, this is not
true in general. Two examples due to Christopher Francisco in which the third power of
the Alexander dual has a number of associated primes which are not associated primes for
the square are the 3-hypergraph with edge ideal
(x x x ,x x x ,x x x ,x x x ,x x x ,x x x ,x x x ,x x x ,x x x ,x x x )
2 4 8 3 5 6 4 7 9 6 8 9 3 4 7 1 2 9 1 4 6 1 8 9 2 5 9 3 4 6
and the complete 2-hypergraph on five vertices.
I would like to thank Dr. Christopher Francisco for his guidance in this project.
1. Tools for Detecting Associated Primes
An m-hypergraph, H, is a vertex set V = {x ,x ,...,x } and an edge set E such that
1 2 n
the elements of E are subsets of V containing m distinct elements. The size of H is said
to be the size of V. A 2-hypergraph is a simple graph (a graph with no loops or multiple
edges).
The edge ideal I of an m-hypergraph H of size n with edge set E and vertex set V is
the ideal generated by the monomials formed by multiplying all the vertices that make up
an edge in a polynomial ring in n variables over a field. The Alexander dual of an edge
ideal I is
I∨ = (x ,x ,...,x ).
\ i1 i2 im
{xi1,xi2,...,xim}∈E
In a 2-hypergraphwith n vertices, a k-cover is a non-zero element of Nn, (a ,a ,...,a ),
1 2 n
such that for every edge {x ,x }, a + a ≥ k. The definition extends nicely to m-
i j i j
hypergraphs: for every edge {x ,x ,...,x } we have a + a + ··· + a ≥ k. Note
i1 i2 im i1 i2 im
that any k-cover is a 0-cover. If a k-cover A can be written as A = B +C where B is
a p-cover and C is a (k − p)-cover, then the k-cover A is said to be reducible. Other-
wise, A is said to be irreducible. A k-cover (a ,a ,...,a ) corresponds to the monomial
1 2 n
xa1xa2···xan. For a k-cover A, we will call the corresponding monomial A. Note that for
1 2 n
2
any monomial Q, there is a k ≥ 0 and a k-cover P such that Q = P. We also have the
observation that for k-covers A and B, A+B = (A)(B).
Lemma 1.1. For all k ≥ 1, the symbolic kth power of the Alexander dual, defined by
(I∨)(k) = (x ,x ,...,x )k
\ i1 i2 im
{xi1,xi2,...,xim}∈E
is generated by the the monomials corresponding to k-covers.
Proof. Since the intersection of powers of monomial ideals is a monomial ideal, it suffices
to show that any monomial in (I∨)(k) corresponds to a k-cover and vice versa. Let M ∈
(I∨)(k) bea monomial. Since M is a monomial, there exists a 0-cover P = (p ,...,p ) such
1 n
thatM = P. SinceM ∈ (I∨)(k),M ∈(x ,x ,...,x )k foralledges{x ,x ,...,x } ∈ E.
i1 i2 im i1 i2 im
Thus for all edges {x ,x ,...,x }∈ E, the sum of the exponents on x ,x ,...,x must
i1 i2 im i1 i2 im
beat least k. Since M = xp1···xpn, we have p +p +···+p ≥ k. Thus P is a k-cover.
1 n i1 i2 im
Reversing the argument, we see that for every k-cover P, P ∈ (I∨)(k). (cid:3)
Thus the following facts are true: The monomials corresponding to irreducible 1-covers
areminimalgeneratorsforI∨,andthemonomialscorrespondingto2-covers aregenerators
for(I∨)(2). Similarlythemonomialscorrespondingto2-coversthatarethesumsof1-covers
are generators of (I∨)2.
By localizing at prime ideals, one can see that (x ,x ,...,x ) is a minimal prime of
i1 i2 im
(I∨)2 if and only if {x ,x ,...,x } ∈ E. If P ,P ,...,P are the other (possibly non-
i1 i2 im 1 2 r
existent) associated primes of (I∨)2 (embedded primes),
(1) (I∨)2 = (x ,x ,...,x )2 ∩Q ∩Q ∩...∩Q
\ i1 i2 im 1 2 r
{xi1,xi2,...,xim}∈E
where Q is P -primary.
i i
Suppose that H is an m-hypergraph. A set of vertices S is an independent set if no
edge contains only vertices in S. The neighborhood of an independent set S, N(S), is
the set of vertices x ∈/ S such that there is an edge {x ,x ,...,x } such that all of the
i i j1 jm
x are in S. Suppose S is an independent set, and T is the neighborhood of S. Define
j
l
A = (a ,...,a ) where a = 0 if x ∈ S, a = 2 if x ∈ T and a = 1 if x ∈/ S ∪T.
s 1 n i i i i i i
Proposition 1.2. Let H be an m-hypergraph.
1. Suppose S is an independent set in H. Then A is a 2-cover.
s
2. Suppose W is a 2-cover, such that W 6= A for all independent sets S. Then there
s
exists a 2-cover X = A for some independent set S and a 0-cover Y such that
s
W = X +Y.
Proof. Let H be an m-hypergraph of size n with vertex set V and edge set E.
1. Suppose S is an independent set in H. Then we have A = (a ,...,a ) where a = 0
s 1 m i
for x ∈ S, a = 2 for x ∈ N(S) and a = 1 for x ∈ V −S−N(S). Let {x ,...,x } bean
i i i i i i1 im
edge in H. We must show that a +a +···+a ≥ 2. Since S is independent, at least
i1 i2 im
one of x ,...,x is not in S. Since a = 0 implies x ∈ S, at least one of a ,a ,...,a
i1 im i i i1 i2 im
is non-zero. If more than one of these is non-zero, we are done. Suppose exactly one of
these, say a , is non-zero. Then we have x ∈S for j 6= k. Thus x ∈ N(S), so a = 2,
ik ij ik ik
and a +a +···+a = 2 ≥ 2. So A is a 2-cover.
i1 i2 im s
2. Suppose W = (w ,w ,...,w ) is a 2-cover, W 6= A for all independent sets S. Let
1 2 n s
S = {x | w = 0}. Then S is an independent set since if there was an edge {x ,...,x }
i i i1 im
3
such that {x ,...,x } ⊂ S, then w +...+w = 0, and this cannot happen since W
i1 im i1 im
is a 2-cover. We will show that W −A is a 0-cover. Let x ∈ N(S). Then there is an
s j
edge {x ,x ,...,x } such that {x ,...,x } ⊂ S. Thus w = 0 for 2 ≤ k ≤ m. Since W
j i2 im i2 im ik
is a 2-cover, w = w +w +···+w ≥ 2. Further, by our definition of S, w ≥ 1 for
j j i2 im i
all x ∈/ S. Thus W −A ∈ Nn. Also, since W 6= A , W −A 6= 0. Since any nonzero
i s s s
element of Nn is a 0-cover, we have W −A is a 0-cover. A is a 2-cover by part 1, so we
s s
have W = A +(W −A ) which proves the proposition. (cid:3)
s s
Theorem 1.3. Let H be an m-hypergraph with edge ideal I. Then (I∨)2 has no non-
minimal associated primes if and only if for every independent set S ⊂ V, A is a sum of
s
two 1-covers.
Proof. Let H be an m-hypergraph with edge ideal I.
Suppose (I∨)2 has no non-minimal associated primes. Then by equation (1),
(I∨)2 = (x ,x ,...,x )2.
\ i1 i2 im
{xi1,xi2,...,xim}∈E
But we also have
(I∨)(2) = (x ,x ,...,x )2,
\ i1 i2 im
{xi1,xi2,...,xim}∈E
so (I∨)2 = (I∨)(2). Further, we know that (I∨)2 is generated by the 2-covers that are the
sums of 1-covers and (I∨)(2) is generated by the 2-covers. Thus the 2-covers are generated
by the 2-covers that are the sums of 1-covers. Thus if A is a monomial corresponding to
a 2-cover, A= QB where B is a 2-cover and Q is another (non-zero) monomial. Thus we
can write A= QB B where B and B are 1-covers. Further, QB must also correspond
1 2 1 2 1
to a 1-cover, so A is the sum of two 1-covers.
Now suppose that for every independent set S, A is a sum of two 1-covers. Let B be
s
a 2-cover. Then by proposition 1.2, B = A for some independent set S, or B = W +A
s s
for some 0-cover W and some independent set S. Whichever is the case, we can say that
A = Q +Q for two 1-covers Q and Q . So B = Q +Q or B = (W +Q )+Q .
s 1 2 1 2 1 2 1 2
Since W +Q is a 1-cover as well, B can be written as the sum of two 1-covers. Thus the
1
set of 2-covers equals the set of 2-covers that can be written as the sum of two 1-covers.
Therefore, (I∨)2 = (I∨)(2). So
(I∨)2 = (x ,x ,...,x )2.
\ i1 i2 im
{xi1,xi2,...,xim}∈E
Since (x ,x ,...,x )2 is (x ,x ,...,x )-primary, we have the irredundant primary
i1 i2 im i1 i2 im
decomposition, so the only associated primes of (I∨)2 are the minimal primes. (cid:3)
Wealsohavethefollowingresult,whichcanbeinvaluableindetectingassociatedprimes:
Theorem 1.4. Let H be an m-hypergraph of size n. Then an ideal P ⊂ K[x ,...,x ] is a
1 n
non-minimal associated prime of (I(H)∨)2 if and only if P = (x ,...,x ) and there exists
i1 iq
a 2-cover C that is not the sum of two 1-covers such that x C corresponds to a 2-cover
ij
that is the sum of two 1-covers for any j ≤ q, and for any W ∈/ (x ,...,x ), WC does not
i1 iq
correspond to a 2-cover that is the sum of two 1-covers.
Proof. Supposetheideal P ⊂ K[x ,...,x ] is a non-minimalassociated primeof (I(H)∨)2.
1 n
Since (I(H)∨)2 is a monomial ideal, P = (x ,...,x ). Then there exists some monomial
i1 iq
Z such that ((I(H)∨)2 : Z) = P. Since Z is a monomial, there exists some 0-cover
4
C = (c ,...,c ) such that Z = C. Note that x Z ∈ (I(H)∨)2 for all j ≤ q. Since (I(H)∨)2
1 n ij
is generated by the monomials corresponding to 2-covers that are the sum of two 1-covers,
x Z corresponds to a 2-cover that is the sum of two 1-covers for all j. Further, since Z
ij
cannot be in (I(H)∨)2, C cannot be the sum of two 1-covers. Since P is non-minimal, P
has height at least m+1. Let R = (r ,...,r ) be the 0-cover such that R = x Z. Then R
1 n i1
is the sum of two 1-covers. Suppose that A= {x ,...,x } is an edge such that x ∈/ A.
a1 am i1
Thenwehaver +···+r ≥ 2. Butfori 6= i , r = c . Thusc +···+c ≥ 2. Suppose
a1 am 1 i i a1 am
that B = {x ,...,x }is an edge whereb = i . Then,since P has height greater than m,
b1 bm 1 1
there exists k such that k = i for some j and k 6= b for all j. Let S = (s ,...,s ) be the
j j 1 n
0-cover such that S =x Z. Then S is the sum of two 1-covers. Thus s +···+s ≥ 2.
k b1 bm
But s = c for all i 6= k. Since k 6= b for all j, c +···+c ≥ 2. Thus C is a 2-cover.
i i j b1 bm
Let W ∈/ P. Then we have WZ ∈/ (I(H)∨)2. Thus WC = WZ does not correspond to a
2-cover that is the sum of two 1-covers.
Now suppose that C = (c ,...,c ) is a 2-cover that is not the sum of two 1-covers,
1 n
and that x C corresponds to a 2-cover that is the sum of two 1-covers for all j ≤ q
ij
and for any W ∈/ (x ,...,x ), WC does not correspond to a 2-cover that is the sum of
i1 iq
two 1-covers. Since (I(H)∨)2 is generated by the monomials that correspond to 2-covers
that are the sum of two 1-covers, C ∈/ (I(H)∨)2 and x C ∈ (I(H)∨)2 for j ≤ q. Thus
ij
(x ,...,x ) ⊂ ((I(H)∨)2 : C). Further, we have for W ∈/ (x ,...,x ), WC ∈/ (I(H)∨)2.
i1 iq i1 iq
Thus (x ,...,x ) = ((I(H)∨)2 : C). Since (x ,...,x ) is a prime ideal, (x ,...,x ) is an
i1 iq i1 iq i1 iq
associated prime of (I(H)∨)2. (cid:3)
The incidence matrix of a hypergraph with v vertices and e edges is the v ×e matrix
with a = 1 if x is a vertex of the jth edge, and a =0 otherwise. If there does not exist
ij i ij
a square submatrix with an odd number of columns such that each row and column has
exactly 2 ones, then the hypergraph is called balanced. It is shown in Proposition 4.11 of
[2] that a balanced hypergraph has no non-minimal associated primes. In the case of a
2-hypergraph (a simple graph), balanced is equivalent to bipartite.
This condition is sufficient to show that a hypergraph has no non-minimal associated
primes, but it is not necessary, as shown by this example, of the 3-hypergraph given by
the edge ideal I = (x x x ,x x x ,x x x ,x x x ), which is not balanced, but has no
1 2 3 3 4 5 5 6 1 2 3 4
non-minimal associated primes.
In the case of 2-hypergraphs, there is a simple necessary and sufficient condition for
(I∨)2 to have no nonminimal associated primes, which we state in Theorem 1.5 below,
with a very simple and direct proof. Theorem 1.5 is also a consequence of facts proven in
Theorem 5.1 of [3].
A 2-hypergraph with vertex set V and edge set E is called bipartite if there exist two
subsets A ⊂ V, B ⊂ V such that V = A∪B, A∩B = ∅ and {x ,x } ∈ E implies either
i j
x ∈ A, x ∈ B or x ∈ B, x ∈ A.
i j i j
Theorem 1.5. SupposeI is the edge ideal of a 2-hypergraph G (a simple graph). Then the
following are equivalent:
1. G is bipartite
2. The 2-cover (1,1,....,1) is a sum of two 1-covers
3. (I∨)2 has no non-minimal associated primes.
Proof. (2 ⇒ 1) Suppose G is a 2-hypergraph of size n. Suppose (1,1,...,1) is the sum of
two 1-covers, Q and P. Then each component of Q and P must be either 1 or 0. Further,
if q = 0, p = 1 and vice versa. Let A = {x | q = 1} and B = {x | p = 1}. Then we
i i i i i i
5
have V = A∪B and A∩B = ∅. Further, since Q and P are both 1-covers, each edge
must have at least one vertex in A and B. Thus G is bipartite.
(1 ⇒ 3) Suppose G is bipartite. Let A ⊂ V, B ⊂ V such that V = A∪B, A∩B = ∅
and {x ,x } ∈ E implies either x ∈ A, x ∈ B or x ∈ B, x ∈ A. Let S be an
i j i j i j
independent set in G. Let A′ = (a ,...,a ) where a = 0 if x ∈ S, a = 1 if x ∈ N(S),
1 n i i i i
a = 1 if x ∈ (V −S −N(S))∩A, and a = 0 if x ∈/ (V −S −N(S))∩A. Similarily, let
i i i i
B′ = (b ,...,b )whereb = 0ifx ∈ S,b = 1ifx ∈ N(S),b = 1ifx ∈ (V−S−N(S))∩B,
1 n i i i i i i
and b = 0 if x ∈/ (V − S − N(S)) ∩ B. Then A + B = A . Let {x ,x } ∈ E. Then
i i s i j
{x ,x } is not a subset of S. Suppose one of x ,x is in S. Then the other is in N(S), so
i j i j
a +a =b +b = 1. Now suppose {x ,x }∩S = ∅. Then {x ,x }⊂ V −S−N(S), and
i j i j i j i j
one of x ,x is in A and the other is in B, so we have either x ∈ (V −S −N(S))∩A,
i j i
x ∈ (V −S −N(S))∩B or vice versa. Either way, a +a = b +b = 1. Thus A′ and
j i j i j
B′ are both 1-covers, so each A is a sum of two 1-covers and by Theorem 1.3, (I∨)2 has
s
no non-minimal associated primes.
(3 ⇒ 2) Suppose (I∨)2 has no non-minimal associated primes. Then by theorem 1.3,
A = (1,1,1,...,1) is the sum of two 1-covers. (cid:3)
∅
Theorem 1.3 gives a generalization of Theorem 1.5 to m-hypergraphs. However, the
combinatorial conditions on Theorem 1.3 are much more complicated on hypergraphs.
For example, the 2-cover (1,1,...,1) can be the sum of two 1-covers when there are non-
minimal associated primes. An example of an edge ideal for such a situation is
I = (x x x ,x x x ,x x x ).
1 2 3 3 4 5 5 6 1
In fact, in m-hypergraphs it is difficult to find an example for which (1,1,...,1) is not the
sum of two 1-covers. However, the complete 3-hypergraph on 10 vertices has this property
(A complete m-hypergraph is an m-hypergraph for which every collection of m vertices is
connected by an edge).
An induced subhypergraph on a subset of the vertex set of an m-hypergraph is the m-
hypergraphformedbythesubsetof thevertex setandthesubsetof theedge setconsisting
of all edges for which all the vertices on the edge are in the subset of the vertex set.
A theorem that provides some insight into the structure of associated primes of (I∨)2
is:
Theorem 1.6. Let H be an m-hypergraph of size n. Then the ideal (x ,...,x ) ⊂
i1 iq
K[x ,...,x ] is an associated prime of (I(H)∨)2 if and only if the ideal (x ,...,x ) ⊂
1 n i1 iq
K[x ,...,x ] is an associated prime of (I(H′)∨)2 where H′ is the induced subhypergraph
i1 iq
on x ,...,x .
i1 iq
Proof. Suppose (x ,...,x )⊂ K[x ,...,x ] is an associated prime of (I(H′)∨)2 where H′
i1 iq i1 iq
is the induced subgraph on x ,...,x . Suppose q = m. Then (x ,...,x ) is a minimal
i1 iq i1 iq
prime,and{x ,...,x }isanedgeinbothH andH′,so(x ,...,x )isanassociated prime
i1 iq i1 iq
in (I(H)∨)2. Now suppose q > m. Then there exists a 2-cover of H′, C = (c ,...,c ),
i1 iq
such that C is not the sum of two 1-covers, but if we add 1 to the i th component of C
k
for any k, we obtain a 2-cover that is the sum of two 1-covers. Let T = (t ,...,t ) be such
1 n
that t = c and t = 2 whenever l 6= i for all j ≤ q. Let {x ,...,x } be an edge in H.
ij ij l j p1 pm
Suppose{x ,...,x }is also anedgein H′. Thent +···+t = c +···+c ≥ 2since
p1 pm p1 pm p1 pm
C is a 2-cover. Suppose at least one of x ,...,x , say x , is not in H′. Then t = 2, so
p1 pm pj pj
t +···+t ≥ 2. ThusT isa2-coverofH. SupposeT = D+E whereD = (d ,...,d )and
p1 pm 1 n
E = (e ,...,e )are1-covers ofH. Thenforalledges{x ,...,x }inH,d +···+d ≥ 1
1 n p1 pm p1 pm
and e +···+e ≥ 1. Specifically, D′ = (d ,...,d ) and E′ = (e ,...,e ) are 1-covers
p1 pm i1 iq i1 iq
6
in H′. Further, E′+D′ = C, which cannot happensince C is not the sum of two 1-covers.
Thus T is not the sum of two 1-covers. Now we must show that if we add 1 to the i th
k
component of T for any k, we obtain a 2-cover that is the sum of two 1-covers, and that if
we add 1 to any other component, we do not. Let F = (f ,...,f ) where f = 1 for some
1 n ik
k, and f = 0 everywhere else. Let F′ = (f ,...,f ). Then there exist two 1-covers of H′,
i i1 iq
A′ = (a′ ,...,a′ ) and B′ = (b′ ,...,b′ ) such that A′ +B′ = F′ +C. Let A = (a ,...,a ),
i1 iq i1 iq 1 n
B = (b ,...,b ) where a = a′ , b = b′ and a = b = 1 everywhere else. Then we have
1 n ij ij ij ij i i
A and B are 1-covers on H and A+B = F +T. Now let F be any 0-cover such that
f = 0 for all k. Then F′ = (0,...,0), and so the same argument that shows that T is not
i
k
the sum of two 1-covers shows that F +T is not the sum of two 1-covers. Thus the ideal
(x ,...,x ) ⊂ K[x ,...,x ] is an associated prime of (I(H)∨)2.
i1 iq 1 n
Now suppose the ideal (x ,...,x ) ⊂ K[x ,...,x ] is an associated prime of (I(H)∨)2.
i1 iq 1 n
Suppose q = m. Then we (x ,...,x ) is minimal and so is an associated prime of
i1 iq
(I(H′)∨)2. Now suppose q > m. Then there exists a 2-cover C = (c ,...,c ) such that
1 n
C is not the sum of two 1-covers, but if we add 1 to the i th component of C for any k,
k
we obtain a 2-cover that is the sum of two 1-covers, and that if we add anything to any
other components, we do not. Let H′ be the induced subhypergraph on {x ,...,x }. Let
i1 iq
C′ = (c′ ,...,c′ ) where c′ = c . Let {x ,...,x } be an edge in H′. Then {x ,...,x }
i1 iq ij ij p1 pm p1 pm
is also an edge in H. Thus c +···+c = c′ +···+c′ ≥ 2. Thus C′ is a 2-cover of
p1 pm p1 pm
H′. SupposeC′ is the sum of two 1-covers. We have C′ = A′+B′ where A′ =(a′ ,...,a′ ),
i1 iq
B′ = (b′ ,...,b′ ) are 1-covers of H′. Let {x ,...,x } be the set of vertices not in H′. Let
i1 iq z1 zr
W = (w ,...,w ) be a 0-cover of H such that W = x2 ···x2 C. Then, for any vertex x
1 n z1 zr j
not in H′, w ≥ 2. Let F = (f ,...,f ) be such that f = 0 if x is in H′, f = 1 otherwise.
j 1 n i i i
Let G = (g ,...,g ) such that g = 0 if x is in H′, g = w −f otherwise. Then we have
1 n i i i i i
for each x not in H′, g ≥f = 1.
i i i
Let A = (a ,...,a ) where a = a′ for 1 ≤ j ≤ q, a = 0 everywhere else. Similarly,
1 n ij ij k
let B = (b ,...,b ) where b = b′ for 1 ≤ j ≤ q, b = 0 everywhere else. Then W =
1 n ij ij k
(A+F)+(B+G). Foreachedge{x ,...,x }notinH′,g +···+g ≥ f +···+f ≥ 1,
p1 pm p1 pm p1 pm
and for each edge {x ,...,x } in H′, a +···+a ≥ 1 and b +···+b ≥ 1. Thus
p1 pm p1 pm p1 pm
W is the sum of two 1-covers. However, this is contrary to the initial supposition about
C, namely that if we add anything to components outside H′, we do not obtain a sum of
two 1-covers. Thus C′ is not the sum of two 1-covers. Now we must show that for any x ,
ij
x C′ corresponds to a sum of two 1-covers. By our initial supposition, x C = ET where
ij ij
E = (e ,...,e )andT = (t ,...,t )are1-covers. LetE′ = (e ,...,e )andT′ = (t ,...,t ).
1 n 1 n i1 iq i1 iq
Then T′ and E′ are 1-covers in H′ Further, T′E′ = x C′. Thus we have (x ,...,x ) is an
ij i1 ij
associated prime of (I(H′)∨)2. (cid:3)
2. m-hypergraphs
In this section we investigate the non-minimal associated primes of (I∨)2 in terms of
properties of the hypergraph. The motivation is the following theorem for 2-hypergraphs
(simple graphs) proven in [1].
Define a cycle as an ordered set C = {x ,x ,...,x } ⊆ V such that for 1 ≤ j < n, x
i1 i2 in ij
and x are connected by an edge on the induced subhypergraph, and x and x are
ij+1 i1 in
connected by an edge. The size of the cycle is the size of C.
7
Acycle C is called inducedif C is nota cycle in thegraph formed by the removal of any
edge from the induced subgraph on C. If a cycle is not induced, it is called non-induced.
The theorem is:
Theorem 2.1. (Theorem 1.1 [1]) Let G be a simple graph (a 2-hypergraph). A prime ideal
P = (x ,...,x ) is an associated prime of (I(G)∨)2 if and only if:
i1 ir
1. P = (x ,x ) and {x ,x } is an edge in G, or
i1 i2 i1 i2
2. r is odd and the set {x ,...,x } is an induced cycle in G for some ordering.
i1 ir
We observe that the primes P in 1 are the minimal primes of (I∨)2 (and (I∨)) and that
a graph G is bipartite if and only if it has no induced cycles of odd size. We thus recover
Theorem 1.5 from Theorem 2.1.
We can easily obtain the following result from Theorem 2.1:
Theorem 2.2. Let P ⊂ K[x ,...,x ] be a monomial prime. Then P is an associated prime
1 n
of (I(G)∨)2 for some 2-hypergraph G of size n if and only if height(P) ≥ 2 or height(P)
is an odd integer greater than or equal to 3.
Inthecaseofanm-hypergraphH,theminimalprimesof(I∨)2aretheprimes(x ,...,x )
i1 im
such that {x ,...,x } is an edge of H, as follows from equation (1). Thus statement 1 of
i1 im
Theorem2.1 generalizes immediately to m-hypergraphs. However, Theorem2.2 andstate-
ment 2 of Theorem 2.1 are only applicable to 2-hypergraphs. In contrast with Theorem
2.2, we have the following theorem for 3-hypergraphs.
Theorem 2.3. Let P ⊂ K[x ,...,x ] be a monomial prime. Then P is an associated prime
1 n
of (I(H)∨)2 for some 3-hypergraph H of size n if and only if height(P) ≥ 3.
The proof of this theorem will be given after Theorem 2.4, which is the essential case
of the proof.
Theorem 2.4. For all n ≥ 3, there exists a 3-hypergraph of size n with edge ideal I such
that there is an associated prime of (I∨)2 of height n.
Proof. We will prove the theorem in 4 cases.
Supposenisanoddnumberoftheform4t+1greaterthan3. LetH bethe3-hypergraph
with vertex set V, edge set E and edge ideal I such that
(2) I = (x x x ,x x x ,x x x ,...,x x x ,x x x ).
1 2 3 3 4 5 5 6 7 n−2 n−1 n n 1 2
Let S = {x ,x ,x ,...,x } where x ∈ S if and only if i ≤ n is even. Clearly, S is an
2 4 6 n−1 i
independent set. Further, N(S) = ∅, so A = (a ,a ,...,a ) where a = 1 if i is odd,
s 1 2 n i
and a = 0 otherwise (we use the notation preceding Proposition 1.2). We will now show
i
that A is not the sum of two 1-covers. Suppose A = Q+P where Q = (q ,...,q ) and
s s 1 n
P = (p ,...,p ) are 1-covers. Without loss of generality, assume q = 1, p = 0. Since
1 n 1 1
a = 0 for i even, q = p = 0 for i even. For all odd k < n, {x ,x ,x } is in E.
i i i k k+1 k+2
Thus if k < n is odd, and q = 1, we have p = 0, q = 0, and p = 1. Similarily, if
k k k+2 k+2
p = 1, we have q = 0, p = 0, and q = 1. We will show by induction that q = 1
k k k+2 k+2 i
if and only if i is of the form 4t+1. Suppose q = 1 if and only if i is of the form 4n+1
i
for all i ≤ 4j + 1. Then we have q = q = 0 since 4j + 2 and 4j + 4 are even,
4j+2 4j+4
and q = 0 since q = 1 and 4j +1 is odd. Further, p = 1. Thus q = 1.
4j+3 4j+1 4j+3 4j+5
4j +5 = 4(j + 1)+ 1. Thus by induction, q = 1 if and only if i = 4t +1 for some t.
i
Further, since A = Q+P, p = 1 if and only if i = 4t+3 for some t. Since n is of the
s i
form 4t+1, p = 0 for i ∈{n,1,2}. But {x ,x ,x } ∈ E. Thus P is not a 1-cover, which
i n 1 2
8
is contradiction. Thus we see that A is not the sum of two 1-covers, and so A is not in
s s
(I∨)2. Now wemustshow that((I∨)2 : A )= (x ,x ,...,x ). Thisis equivalent toproving
s 1 2 n
that x A ∈ (I∨)2 for all i ≤ n. Let k ≤ n. Suppose k is odd. Then let Q = (q ,...,q )
i s 1 n
where q = 1 and q = 1 for all odd i of the form 4t+1 less than or equal to k and q = 1
1 i i
for all odd i of the form 4t+3 greater than or equal to k, and q = 0 everywhere else.
i
Let P = (p ,...,p ) where p = 1 for all odd i of the form 4t+3 less than or equal to k,
1 n i
p = 1 for all odd i of the form 4t+1 greater than or equal to k , and p = 0 everywhere
i i
else. Then it is clear that x A = Q+P. For all odd i, either q = 1 and p = 1 or vice
k s i i−2
versa. Also, p = 1, so Q and P are 1-covers. Suppose k is even, and of the form 4t. Let
n
Q = (q ,...,q ) where q = 1, q = 1 and q = 1 for all odd i of the form 4t+1 less than
1 n k 1 i
or equal to k and q = 1 for all odd i of the form 4t+3 greater than or equal to k, and
i
q = 0 everywhere else. Let P = (p ,...,p ) where p = 1 for all odd i of the form 4t+3
i 1 n i
less than or equal to k, p = 1 for all odd i of the form 4t+1 greater than or equal to k ,
i
and p = 0 everywhereelse. Then itis clear that x A = Q+P. Thenfor i 6= k+1, either
i k s
q = 1 and p = 1 or vice versa. Further, for i = k+1, p = p = 1, but q = 1. Thus
i i−2 i i−2 k
Q and P are 1-covers. Finally, suppose k is of the form 4t+2. Let Q and P be as in the
previous cases, with the difference that q =0 and p = 1. Then once again, Q and P are
k k
1-covers and x A = Q+P. Thus x A ∈ (I∨)2. Therefore (x ,x ,...,x ) = ((I∨)2 : A )
k s k s 1 2 n s
is an associated prime of (I∨)2.
Now suppose n is an odd number of the form 4t+3 greater than 3. Let
(3) I = (x x x ,x x x ,x x x ,...,x x x ,x x x ,x x x ).
1 2 3 3 4 5 5 6 7 n−2 n−1 n n 1 2 4 5 6
Let S = {x ,x ,x ,x ,x ,...,x }, so x ∈ S if and only if i = 5, i = 2, or i is an
2 5 8 10 12 n−1 i
even number greater than or equal to 8 and less than or equal to n−1. S is clearly an
independent set, and N(S) = ∅. We will show that A = (a ,...,a ) where a = 0 if
s 1 n i
x ∈ S and a = 1 if x ∈/ S is not the sum of two 1-covers. Suppose A = Q+P where
i i i s
Q = (q ,...,q ) and P = (p ,...,p ) are 1-covers. Without loss of generailty, suppose
1 n 1 n
q = 1. Then we must have p = 1, q = 1, p = 1, and q = 1 since Q and P are
1 3 4 6 7
1-covers. Now we may use the argument in the proof of part 1 to say that if i is greater
than 7, q = 1 if and only if i is of the form 4t+3, and p = 1 if and only if i is of the
i i
form 4t+1. However, we now have p = 0 for i ∈ {n,1,2} since n is of the form 4t+3.
i
But {x ,x ,x } ∈ E, so P is not a 1-cover. Thus A cannot be written as the sum of
n 1 2 s
two 1-covers. Thus A ∈/ (I∨)2. Now we must show that x A ∈ (I∨)2 for 1 ≤ i ≤ n,
s i s
which once again means we must show that x A corresponds to a sum of two 1-covers.
i s
Let Q and P be as just defined, so Q+P = A , but P is not a 1-cover. Let k ≤ n. Let
s
F = (f ,...,f ), G = (g ,...,g ), where f = q for i < k, f = p for i > k, and g = p for
1 n 1 n i i i i i i
i < k and g = q for i > k. and f = g = 1 if a = 1, f = 1, g = 0 if p = 1, and
i i k k k k k k−1
vice versa if q = 1. Then we will have F +G = x A and F and G are 1-covers. Thus
k−1 k s
x A ∈ (I∨)2, and so (x ,...,x ) is an associated prime of (I∨)2.
k s 1 n
Now suppose n is an even number, and n = 4z+2 for some z ≥ 1. Let
(4) I = (x x x ,x x x ,x x x ,...,x x x ,x x x ).
1 2 3 3 4 5 5 6 7 n−3 n−2 n−1 n−1 n 1
Let S = {x ,x ,...,x }, wherex ∈ S if andonly ifi ≤ nis even. S is clearly independent,
2 4 n i
and N(S) = ∅. We will show that A = (a ,...,a ) is not the sum of two 1-covers. Note
s 1 n
that all the components of A are either 1 or 0. Further, note that any vertex in the
s
hypergraph is included in at most 2 edges. Thus if W = (w ,...,w ) is some element of
1 n
Nn whose components are either 0 or 1 and has q ones, then at most 2q edges {x ,x ,x }
i j k
havethepropertyw +w +w ≥ 1. ThusinorderforanelementofNn whosecomponents
i j k
9
are all 0 or 1 to be a 1-cover, at least n components must be 1 since our hypergraph has
4
n edges. Since n = 4z +2, this means that a 1-cover must have at least z +1 non-zero
2
components. A has n = 2z + 1 components whose value is 1. Thus if A = Q + P
s 2 s
where Q is a 1-cover, P can have at most z non-zero components, and so P is not a
1-cover. Therefore A is not a sum of 1-covers, and A ∈/ (I∨)2. Now we must show that
s S
x A ∈ (I∨)2 for all k ≤ n. Suppose k odd. Let Q = (q ,...,q ), P = (p ,...,p ) where
k s 1 n 1 n
q = 1 for odd i < k of the form 4t+1, q = 1 for odd i > k of the form 4t+3, q = 1
i i k
and all other q = 0. Similarily, let p = 1 for odd i < k of the form 4t+3, p = 1 for odd
i i i
i > k of the form 4t+1, p = 1 and all other p = 0. Then it is clear that Q+P = x A .
k i k s
Further, Q and P are 1-covers. Suppose k is even. Let Q and P be as before, with the
alteration that q = 1 if only if k is of the form 4t, and p = 1 if only if k is of the form
k k
4t+2. Then once again, Q+P = x A and Q and P are 1-covers. Thus (x ,...x ) is an
k s 1 n
associated prime of (I∨)2.
Now suppose n is an even number, and n = 4z for some z ≥ 1. Let
(5) I = (x x x ,x x x ,x x x ,...,x x x ,x x x ,x x x ).
1 2 3 3 4 5 5 6 7 n−3 n−2 n−1 n−1 n 1 2 3 4
Let S = {x ,x ,x ,...,x }, so x ∈ S if and only if i = 3 or i is an even number greater
3 6 8 n i
than or equal to 6 and less than or equal to n. S is clearly independent, and N(S) = ∅.
Again, we will show A = (a ,...,a ) is not a sum of 1-covers. SupposeA = Q+P where
s 1 n s
Q = (q ,...,q ) and P = (p ,...,p ) are 1-covers. Without loss of generality, suppose
1 n 1 n
q = 1. Then we must have p = 1, q = 1 and p = 1 in order for Q and P to be 1-covers.
1 2 4 5
Now we may use an argument as in the previous cases to show that for i larger than 5,
q = 1 if and only if i is of the form 4n+3 and p = 1 if and only if i is of the form 4n+1.
i i
However, p = 0 for i ∈ {n−1,n,1} since n = 4z. Since {x ,x ,x } ∈ E, P is not a
i n−1 n 1
1-cover. Thus A is not a sum of two 1-covers. Now we show that x A ∈ (I∨)2 for all
s k s
k ≤ n. Let F = (f ,...,f ), G = (g ,...,g ) be defined as follows. Let F be such that
1 n 1 n
f = q for i < k, f = p for i > k, f = 1 if a = 1 or p = 1 and f = 0 otherwise.
i i i i k k k−1 k
Similarily, let G be such that g = p for i < k, g = q for i > k, g = 1 if a = 1 or
i i i i k k
q = 1 and g = 0 otherwise. Then F and G are 1-covers and F +G = x A . Thus
k−1 k k s
(x ,...,x ) is an associated prime of (I∨)2. (cid:3)
1 n
Now we will prove Theorem 2.3. Suppose a monomial prime P ⊂ K[x ,...,x ] is an as-
1 n
sociated primeof(I(H)∨)2 forsome3-hypergraphH. Sincetheminimalassociated primes
of(I(H)∨)2 areofheight3,height(P) ≥ 3. SupposeP ⊂ K[x ,...,x ]isamonomialprime
1 n
with height(P) ≥ 3. Then P = (x ,...,x ) for some 1 ≤ i < ··· < i ≤ n. By Theo-
i1 ir 1 r
rem 2.4, there exists a 3-hypergraph H of size r such that the ideal Q = (x ,...,x ) ⊂
i1 ir
K[x ,...,x ] is an associated prime of (I(H)∨)2. Let H′ be the 3-hypergraph with vertex
i1 ir
set {x ,...,x } and the same edge set as H. Then, becausethey have thesame generators,
1 n
(I(H′)∨)2 = (I(H)∨)2K[x ,...,x ]. Further, P = K[x ,...,x ]Q, so P is an associated
1 n 1 n
prime of (I(H′)∨)2. Thus we have a 3-hypergraph of size n such that P is an associated
prime of the square of the Alexander dual of the edge ideal.
A hypergraph is connected if for any two vertices x and x , there exists an ordered set
i j
of vertices {x ,...,x } such that x = x , x = x and x is connected by an edge to
p1 pq p1 i pq j pk
x for all k < q.
pk+1
Corollary 2.5. For any integers n and q, n≥ q ≥ 3, there exists a connected 3-hypergraph
of size n such that the square of the Alexander dual of the edge ideal has an associated
prime of height q.
10
